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A355585
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T(j,k) are the numerators s in the representation R = s/t + (2*sqrt(3)/Pi)*u/v of the resistance between two nodes separated by the distance (j,k) in an infinite triangular lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= floor(j/2) is an irregular triangle read by rows.
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12
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0, 1, 8, -2, 27, -5, 928, -70, 16, 11249, -2671, 123, 46872, -34354, 5992, -438, 1792225, -445535, 28075, -10303, 23152256, -5824226, 1168304, -178754, 38336, 100685835, -25547957, 5343755, -885717, 101355, 3970817992, -338056246, 72962904, -12914726, 1825464, -386166
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OFFSET
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0,3
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COMMENTS
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The distance vector (j,k) is defined in an oblique coordinate system with an angle of 120 degrees between the axes, see e.g. A307012.
Atkinson and Steenwijk (1999) (see links in A211074) provided a generalization of the method used to calculate the resistance between two arbitrary nodes in an infinite square lattice of one-ohm resistors to infinite triangular lattices. Similar to the square lattice, the integral describing the resistance distance between nodes can exactly be represented by an expression of the form given in the name of this sequence with integer coefficients. Atkinson and Steenwijk, page 489, provided results for j <= 3 found by evaluation of the integral (17) (given below) and application of Mathematica's "Simplify" function.
R(j,k) = (1/Pi) * Integral_{y=0..Pi/2} (1 - exp(-|j-k|*x)*cos((j+k)*y)) / (sinh(x)*cos(y)) dy, with x = arccosh(2/cos(y)-cos(y)).
It would be useful to know whether, since the publication cited, a recurrence analogous to that known for the square lattice (used in A355565) for determining the coefficients has also been found for the triangular lattice.
The results in this sequence were found by systematic parameter variation of u and v and continued fraction expansion of the difference from the exact value of the integral for the resistance distance to determine s/t.
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REFERENCES
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See A211074 for more references and links (with alternatives).
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LINKS
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Hugo Pfoertner, PARI program for inverse problem, (2022). Finds the grid point [x,y] that leads to the best approximation of a given resistance distance R (ohms) between [0,0] and [x,y].
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FORMULA
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EXAMPLE
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The triangle begins:
0;
1;
8, -2;
27, -5;
928, -70, 16;
11249, -2671, 123;
46872, -34354, 5992, -438;
1792225, -445535, 28075, -10303;
23152256, -5824226, 1168304, -178754, 38336;
100685835, -25547957, 5343755, -885717, 101355;
. The combined triangles used to calculate the resistances are:
\ j 0 | 1 |
k\---------- s/t ----------- u/v -|----------- s/t ----------- u/v -|
0| 0/1 0/ 1 | . . |
1| 1/3 0/ 1 | . . |
2| 8/3 -2/ 1 | -2/3 1/ 1 |
3| 27/1 -24/ 1 | -5/1 5/ 1 |
4| 928/3 -280/ 1 | -70/1 64/ 1 |
5| 11249/3 -3400/ 1 | -2671/3 808/ 1 |
6| 46872/1 -212538/ 5 | -34354/3 51929/ 5 |
7| 1792225/3 -2708944/ 5 | -445535/3 673429/ 5 |
8| 23152256/3 -244962336/35 | -5824226/3 61623224/35 |
9| 100685835/1 -3195918288/35 | -25547957/1 810930216/35 |
10| 3970817992/3 -42013225014/35 | -338056246/1 2146081719/ 7 |
11| 52514317745/3 -111125508824/ 7 | -13481564911/3 142641647567/35 |
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continued
\ j 2 | 3 |
k\-------- s/t ---------- u/v -|--------- s/t -------- u/v -|
4| 16/1 -14/ 1 | . . |
5| 123/1 -111/ 1 | . . |
6| 5992/3 -9054/ 5 | -438/1 1989/5 |
7| 28075/1 -127303/ 5 | -10303/3 15576/5 |
8| 1168304/3 -12361214/35 | -178754/3 1891328/35 |
9| 5343755/1 -169618717/35 | -885717/1 28113999/35 |
10| 72962904/1 -2315951182/35 | -12914726/1 81986531/ 7 |
11| 993810715/1 -31545031729/35 | -184858117/1 5867671888/35 |
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continued
\ j 4 | 5 |
k\------- s/t -------- u/v -|------- s/t ------- u/v -|
8| 38336/3 -405592/35 | . . |
9| 101355/1 -3217136/35 | . . |
10| 1825464/1 -57942922/35 | -386166/1 12257507/35 |
11| 28123355/1 -892677136/35 | -3085317/1 97932579/35 |
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Using the terms for (j,k) = (10,5) with {s, t, u, v} = {-386166, 1, 12257507, 35} the resistance is R = T(10,5)/A355586(10,5) + (2*sqrt(3)/Pi) * A355587(10,5)/A355588(10,5) = -386166/1 + (2*sqrt(3)/Pi)*12257507/35 = 0.731139136228538824636... . This equals the integral for the resistance distance R(j,k) after substitution of j=10 and k=5.
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PROG
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(PARI) Rtri(n, p)={my(alphat(beta)=acosh(2/cos(beta)-cos(beta))); intnum (beta=0, Pi/2, (1 - exp (-abs(n-p) * alphat(beta))*cos((n+p)*beta)) / (cos(beta)*sinh(alphat(beta)))) / Pi};
searchr (target, maxn=1000000, maxd=10, maxrat=1000, minn=0, mind=1) = {my (Rcons=2*sqrt(3)/Pi, delta=oo); for (d=mind, maxd, my(PP=Rcons/d); for (nn=minn, maxn, foreach ([-nn, nn], n, my (P=PP*n, T=target-P, Q = bestappr(T, maxrat), D=abs(target-P-Q)); if(D<delta, delta=D; if (delta < 10^(-getlocalprec()*5\6), return ([numerator(Q), denominator(Q), n, d]) )))))};
\\ j=6 takes a while; prints [j, k][s, t, u, v]
for (j=0, 6, for (k=0, j\2, my(R=Rtri(j, k)); S=searchr(R, 500000); print([j, k], S)));
(PARI) \\ Alternative method using a recurrence; calculates triangle of s/t
jk(j, k) = {my(jj=j, kk=k); if(k<1, jj=j-k+1; kk=2-k); my(km=(jj+1)/2); if(kk>km, kk=2*km-kk); [jj, kk]};
D(n) = subst(pollegendre(n), 'x, 7);
ST(nend) = {my(nmax=nend+1, N=matrix(nmax, (nmax+1)\2)); for (n=2, nmax, N[n, 1]=(1/3) * sum(k=0, n-2, D(k))); for (n=3, nmax, N[n, 2] = (1/2)*(6*N[n-1, 1] - 2*N[jk(n-1, 2)[1], jk(n-1, 2)[2]] - N[n-2, 1] - N[n, 1])); for (n=5, nmax, for (m=3, (n+1)\2, N[n, m] = 6*N[jk(n-1, m-1)[1], jk(n-1, m-1)[2]] - N[jk(n-1, m)[1], jk(n-1, m)[2]] - N[jk(n-2, m-1)[1], jk(n-2, m-1)[2]] - N[jk(n-2, m-2)[1], jk(n-2, m-2)[2]] - N[jk(n-1, m-2)[1], jk(n-1, m-2)[2]] - N[jk(n, m-1)[1], jk(n, m-1)[2]] )); N};
ST(11)
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CROSSREFS
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A355586 are the corresponding denominators t.
Cf. A307012 (discussion of oblique coordinate system).
Cf. A084768 (when divided by 3 apparently gives the difference between successive values of s/t in column 0).
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KEYWORD
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tabf,frac,sign
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AUTHOR
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STATUS
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approved
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